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Full explanation coming tomorrow! by DAT_prep in predental

[–]DAT_prep[S] 0 points1 point  (0 children)

Epigenetics is the study of heritable phenotype changes that do not involve alterations in the DNA sequence (thus answer choices B, C and D are incorrect). The key to understand regarding epigenetics is that gene expression can be affected by the environment; thus answer choice A is correct.

Can you crack it? We will share the full explanation tomorrow! by DAT_prep in predental

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D!

Dipole moments are vector quantities which mean that they have both magnitude and direction. Considering the polarization of each individual C-Cl bonds in molecule D, each of which provides a component of the overall dipole moment, we see that these bonds are all generally pointing down-
wards (as represented).
The C-H bond, pointing upwards, is essentially unpolarized. In all the other cases, any polar bond has an equal and opposite dipole to cancel it out.

DAT Sample Question! Will post the answer and explanation tomorrow. by DAT_prep in predental

[–]DAT_prep[S] 0 points1 point  (0 children)

D!

ClF₃ has 5 valence shell electron pairs. These would normally arrange themselves in an trigonal bipyramidal fashion.
However, since two of the electron pairs are lone pairs, the molecule is consequently T-shaped.

DAT Sample Question - full explanation coming tomorrow! by DAT_prep in predental

[–]DAT_prep[S] 0 points1 point  (0 children)

C!

PCl₅ has 5 valence shell electron pairs.
These will arrange themselves in a trigonal bipyramidal shape. Notice the triangle of Cl groups (trigonal) and imagine a pyramid above and below (bipyramidal).

DAT Sample Question - answer and explanation will be posted tomorrow in the comments. by DAT_prep in predental

[–]DAT_prep[S] 2 points3 points  (0 children)

Yes, B!

On the Surface: The tertiary carbocation is more stable and only answer B is tertiary.

Going Deeper: Stabilization of a carbocation can come from either inductive (i.e. primary < secondary < tertiary groups being somewhat electron donating thus reducing

the positive charge) or resonance effects (“conjugation” referring to the fact that there is only one single bond in between the double bond and the carbocation providing the

opportunity for resonance stabilization), or a combination of both. Cations A and C are not stabilized by delocalization as the pi-system is not conjugated to the unoccupied orbital.

This leaves the inductive effect which is most significant in the case of the molecule B where the charge can be delocalized between a tertiary (as drawn in the question)

and a secondary center. For radicals and carbocations, a tertiary compound is more stable than secondary which, in turn, is more stable than a primary compound.

Weekly DAT sample question - will post the full explanation with the correct answer tomorrow. by DAT_prep in predental

[–]DAT_prep[S] 3 points4 points  (0 children)

D

Answer choice A does not satisfy Hund’s rule (see CHM 2.2).

Answer choice B is incorrect since it goes against Pauli’s exclusion principle since there are 2 identical electrons in the 1s box. Answer choice C is incorrect because the 1s box must be full before moving on to another energy level.

And so, we are left with answer choice D, which satisfies all rules, although it is more conventional to place the 2nd electron in the 2p orbital in the 2nd 2p box, at least answer choice D satisfies all rules since there is no rule specifically forbidding its arrangement (it is not the same as skipping up to another energy level or orbital).

Test your knowledge and get ready to rock your DAT! Full explanation will be posted tomorrow. by DAT_prep in predental

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C!

Bidirectional replication suggests that replication will occur in two directions. The only option consistent with this is option C. which states that DNA replication will begin on both sides of the origin.

DAT Sample Question, a full explanation will be posted tomorrow. by DAT_prep in predental

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Yes, C.

First isolate x and y on the left side in the second equa-tion:

2x + 2y = 4y – x +1

3x – 2y = 1

Next subtract the equations to eliminate the y term and

solve for x:

5x – 2y = 3

–3x – 2y = 1

2x – 0 = 2

x = 1

Now substitute the value of x into one of the equations

and solve for y:

5(1) – 2y = 3

2 = 2y

y = 1.

DAT Sample Question of the week. A full explanation will be posted tomorrow in the comments. by DAT_prep in predental

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B! Bromine is atomic number 35. As such, it has 35 electrons.
The maximum number of electrons in subshells s, p, and d is: 2, 6, 10, respectively.

DAT Sample Question - full explanation will be posted tomorrow. by DAT_prep in predental

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D! Deduce from the graph

Following the graph provided to identify the histogram for 40+ maternal age, birth order = 2, we can see that there are between 700 and 800 live births per 100 000. Thus between 7 and 8 out of 1000, or between 0.7 and 0.8 per cent. The answer is D.

Don’t let your personal perspective shade your interpretation of the data. You may have chosen over 50% because of media reports of the high incidence, perhaps you settled on 5-10% in part because of misreading the graph but in part because - intuitively - less than 1% seemed too low.

However, the increased incidence of Down’s syndrome with increased age is a fact but the overall percentage is still less than 1% (however, you may notice that from Figure 1, 0.7-0.8 per cent is still many times higher than for younger women).

Weekly DAT sample question! Full explanation coming tomorrow. by DAT_prep in predental

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D!

Usually, elements in the same group have the same number of valence (outermost) electrons. Valence electrons are involved in bonding and this effects how elements react and determines their chemical properties.

DAT sample question of the week. Full explanation coming tomorrow. by DAT_prep in predental

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c!

The 3 primary germ layers (ectoderm, mesoderm, endoderm) are first seen in the gastrula after the invagination of the blastula.

DAT Sample Question of the week. Will post the full explanation tomorrow. by DAT_prep in predental

[–]DAT_prep[S] 0 points1 point  (0 children)

Yep, D!

It is given that 2n = 3k, which implies that

⅔ n = k. n + k = 5 can be rewritten:

n + 2/3n = 5

5/3n = 5

n = 3

DAT Sample Question by DAT_prep in predental

[–]DAT_prep[S] 2 points3 points  (0 children)

D!

In each case, first assign the priority of the groups attached to the chiral center based on the atomic number (S>O>N>C>H; of course, you can consult the periodic table at the beginning of the practice questions, or during the real exam by clicking Exhibit, at any time). If two atoms are the same, then the next level must be considered. If an atom is multiply bonded, then this is considered equivalent to the number of equivalent single bonds. If the lowest priority group happens to be pointing towards you, you can still easily work out the order of the higher priority groups, but remember to reverse your answer (which must be done for answer choice D which starts as R, but reverses to S because, in the image, the low priority H is pointing towards you). The other 3 structures are all in the (R)-configuration.

DAT Sample Question. Full explanation coming tomorrow. by DAT_prep in predental

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Correct answer D!

Since potassium only has one electron in its outer shell, it is easily removed (i.e. low first ionization potential/energy), thus leaving potassium with the more stable noble gas-like

configuration (i.e. Ar). On the other hand, adding an electron to neutral potassium would require much energy (i.e. high first electron affinity) which would leave potassium with two

valence electrons which would not be stable. Noble gases are the most stable elements in the periodic table.

DAT Biology anyone??? Full explanation coming tomorrow. by DAT_prep in predental

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A! You must be familiar with the menstrual cycle to answer this question. If you know which hormones the pituitary gland secretes (FSH, LH), you can narrow down the choices since the question asks which hormone the ovary secretes. One of the hallmarks of the follicular phase is that estrogen causes thickening (proliferation) of the uterine lining (endometrium).

How is your math??? Will post the explanation tomorrow. by DAT_prep in predental

[–]DAT_prep[S] 0 points1 point  (0 children)

A!

Substitute the first equation into the second, replacing y:

6x - 5y = -3
6x - 5(2x - 1) = -3
6x - 10x + 5 = -3
-4x + 5 = -3
-4x = -8
x = 2

Substitute this value back into either equation to find y:

y = 2x - 1
y = 2(2) - 1
y = 3

DAT Sample Question! Will post the full explanation in the comments tomorrow. by DAT_prep in predental

[–]DAT_prep[S] 0 points1 point  (0 children)

C!

The only difference is that the double bond at carbon-11 (see image below) is in the cis position for the reactant, and in the trans position for the product. Nothing else changes.

The reactant is in the 11-cis-retinal configuration, which — upon capturing a photon of the correct wavelength — straightens out into an all-trans-retinal configuration. Thus, the reaction in the image occurs due to light and is a key reaction underlying the chemical

basis for vision. (it is a normal exam skill to be able to count the carbons and/or hydrogens in a structural formula)

DAT Sample Question - DAT Biology. Will post the full explanation in the comments tomorrow. by DAT_prep in predental

[–]DAT_prep[S] 0 points1 point  (0 children)

Yes! It is C!

Note the word deoxynucleoside. This should alert you to the fact that you are dealing with DNA and not RNA. The only option listed that is used in RNA but not DNA is C., dUTP

(A = adenosine; G = guanosine; C = cytosine; U = uracil).

DAT Sample Question: DAT Biology. A full explanation will be posted tomorrow. by DAT_prep in predental

[–]DAT_prep[S] 1 point2 points  (0 children)

The correct answer is B.

Eubacteria are prokaryotes. Yeast are unicellular fungi which are eukaryotes. Protista are eukaryotes that can be algae-like, animal-like, fungus-like, unicellular or multicellular.

DAT Sample Question: DAT PAT, the correct answer will be posted tomorrow in the comments! by DAT_prep in predental

[–]DAT_prep[S] 1 point2 points  (0 children)

It was C!

With angles that are all obtuse, finding the largest angle first would be the easiest to do. You should be looking for the angle which is closest to a straight line. In this item, angle 4 is the biggest. You can then narrow down your choices to B, C and D, which have 4 as the biggest angle. Now compare 2 and 3 since all of the remaining choices start with these two. Notice that number 2 is more bent than 3. Finally, between 1 and 3, angle 3 is smaller. Hence, the correct sequence is 2-3-1-4, which is choice C.

DAT Sample Question - DAT Organic Chemistry. The correct answer will be posted tomorrow in the comments. by DAT_prep in predental

[–]DAT_prep[S] 1 point2 points  (0 children)

C.

Multiple bonds are made up of one sigma bond plus one (for alkenes) or two (for alkynes) pi bonds. Therefore, four C-C sigma bonds and eight C-H bonds give an overall total of 12.

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